# proving a trig identity

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prove the identity

[cos2A + sin2A - 1]/[cos2A - sin2A + 1] = tanA

i've been stuck on this one and i'm not sure where to start, some help would be great :-)

[cos2A + sin2A - 1]/[cos2A - sin2A + 1] = tanA

i've been stuck on this one and i'm not sure where to start, some help would be great :-)

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#2

I would start by turning cos2A and sin2A into their cosA and sinA counterparts

Assuming you know the formulas?

Assuming you know the formulas?

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(Original post by

I would start by turning cos2A and sin2A into their cosA and sinA counterparts

Assuming you know the formulas?

**JustSomeGuy:/**)I would start by turning cos2A and sin2A into their cosA and sinA counterparts

Assuming you know the formulas?

that would be:

[cos^2A - sin^2A +2sinAcosA - 1] / [cos^2 - sin^2A - 2sinAcosA + 1]

Last edited by SashaNilssen; 1 month ago

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#4

(Original post by

yeah, the sin2A and cos2A identities.

that would be:

[cos^2A - sin^2A +2sinAcosA - 1] / [cos^2 - sin^2A - 2sinAcosA + 1]

**SashaNilssen**)yeah, the sin2A and cos2A identities.

that would be:

[cos^2A - sin^2A +2sinAcosA - 1] / [cos^2 - sin^2A - 2sinAcosA + 1]

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(Original post by

Now I would divide the top and bottom expressions by cos^2A. It seems like a really random thing to do but you'll notice that now everything is in terms of tanA and secA, which is helpful because there's another formula that links them 2.

**JustSomeGuy:/**)Now I would divide the top and bottom expressions by cos^2A. It seems like a really random thing to do but you'll notice that now everything is in terms of tanA and secA, which is helpful because there's another formula that links them 2.

[cos^2A - sin^2A + sinA - 1] / [cos^2A- sin^A - sinA + 1]

hm, i don't see how this can be written with tanA and secA... did i do something wrong here?

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#6

(Original post by

okay so after doing that i get

[cos^2A - sin^2A + sinA - 1] / [cos^2A- sin^A - sinA + 1]

hm, i don't see how this can be written with tanA and secA... did i do something wrong here?

**SashaNilssen**)okay so after doing that i get

[cos^2A - sin^2A + sinA - 1] / [cos^2A- sin^A - sinA + 1]

hm, i don't see how this can be written with tanA and secA... did i do something wrong here?

(cos^2A/cos^2A) - (sin^2A/cos^2A) + (sinAcosA/cos^A) - (1/cos^2A) and then simplify accordingly. You would do the same with the bottom as well.

The reason why is because if you divide the top AND bottom by cos^A it doesn't change the expression... it's like multiplying the fraction by 1 because you've done the same thing to the top and bottom so you have the same expression as before just it looks different so you can manipulate it better to get the answer you want.

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(Original post by

You didn't divide the WHOLE expression by cos^2A. So for example the top which is cos^2A-sin^2A+sinA-1 would become:

(cos^2A/cos^2A) - (sin^2A/cos^2A) + (sinAcosA/cos^A) - (1/cos^2A) and then simplify accordingly. You would do the same with the bottom as well.

The reason why is because if you divide the top AND bottom by cos^A it doesn't change the expression... it's like multiplying the fraction by 1 because you've done the same thing to the top and bottom so you have the same expression as before just it looks different so you can manipulate it better to get the answer you want.

**JustSomeGuy:/**)You didn't divide the WHOLE expression by cos^2A. So for example the top which is cos^2A-sin^2A+sinA-1 would become:

(cos^2A/cos^2A) - (sin^2A/cos^2A) + (sinAcosA/cos^A) - (1/cos^2A) and then simplify accordingly. You would do the same with the bottom as well.

The reason why is because if you divide the top AND bottom by cos^A it doesn't change the expression... it's like multiplying the fraction by 1 because you've done the same thing to the top and bottom so you have the same expression as before just it looks different so you can manipulate it better to get the answer you want.

i had [cos^2A - sin^2A +2sinAcosA - 1] / [cos^2 - sin^2A - 2sinAcosA + 1]

and then [cos^2A - sin^2A +sinAcosA - 1] / cos^2A

what happens to the original numerator?

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#8

(Original post by

okokok thats great, i simplified and i got to the final answer of tanA, but i have one more concern, what happens to the other numerator that was there before i divided by cos^2A?

i had [cos^2A - sin^2A +2sinAcosA - 1] / [cos^2 - sin^2A - 2sinAcosA + 1]

and then [cos^2A - sin^2A +sinAcosA - 1] / cos^2A

what happens to the original numerator?

**SashaNilssen**)okokok thats great, i simplified and i got to the final answer of tanA, but i have one more concern, what happens to the other numerator that was there before i divided by cos^2A?

i had [cos^2A - sin^2A +2sinAcosA - 1] / [cos^2 - sin^2A - 2sinAcosA + 1]

and then [cos^2A - sin^2A +sinAcosA - 1] / cos^2A

what happens to the original numerator?

Last edited by JustSomeGuy:/; 1 month ago

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(Original post by

Would you mind sending your working between how you got from the first thing to the second? If it helps you can just take a picture and send it from your phone.

**JustSomeGuy:/**)Would you mind sending your working between how you got from the first thing to the second? If it helps you can just take a picture and send it from your phone.

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#10

In case I explained it bad this is what I meant by divide by cos^2A

Edit: sorry that it's sideways I didn't realize it would turn out that way lmao

Edit: sorry that it's sideways I didn't realize it would turn out that way lmao

Last edited by JustSomeGuy:/; 1 month ago

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(Original post by

In case I explained it bad this is what I meant by divide by cos^2A

Edit: sorry that it's sideways I didn't realize it would turn out that way lmao

**JustSomeGuy:/**)In case I explained it bad this is what I meant by divide by cos^2A

Edit: sorry that it's sideways I didn't realize it would turn out that way lmao

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#12

(Original post by

ohhh okay i wont bother getting the image onto my pc, i understand it much better now thank you, i had the wrong idea

**SashaNilssen**)ohhh okay i wont bother getting the image onto my pc, i understand it much better now thank you, i had the wrong idea

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#13

**SashaNilssen**)

okay so after doing that i get

[cos^2A - sin^2A + sinA - 1] / [cos^2A- sin^A - sinA + 1]

hm, i don't see how this can be written with tanA and secA... did i do something wrong here?

the one you chose is not the best for this question.

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(Original post by

Ah right cool my bad for the explanation then lol just tell me if you get it in the end or not and I'll be sure to help you out

**JustSomeGuy:/**)Ah right cool my bad for the explanation then lol just tell me if you get it in the end or not and I'll be sure to help you out

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(Original post by

remember there are 3 versions of Cos(2a).

the one you chose is not the best for this question.

**the bear**)remember there are 3 versions of Cos(2a).

the one you chose is not the best for this question.

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#16

(Original post by

wait no i have been mistaken yet again oops, i hit a wall

**SashaNilssen**)wait no i have been mistaken yet again oops, i hit a wall

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#17

Referring to what "the bear" said. If you used cos2A = 1-2sin^2A for the top and cos2A=2cos^2A-1 for the bottom it would also give you the answer and would be seen as easier since you don't have the do the whole diving by cos^A thing however knowing which ones to use and where can be just as hard.

Last edited by JustSomeGuy:/; 1 month ago

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(Original post by

My final tip to you is that sec^A = tan^A + 1 which is definitely a formula you should learn but you can derive it from sin^A+cos^A=1 if you divide both sides by cos^2A

**JustSomeGuy:/**)My final tip to you is that sec^A = tan^A + 1 which is definitely a formula you should learn but you can derive it from sin^A+cos^A=1 if you divide both sides by cos^2A

thank you very much for your patience :-)

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#19

(Original post by

yeah i know that one, i probably forgot, i did that and simplified and i got my answer.

thank you very much for your patience :-)

**SashaNilssen**)yeah i know that one, i probably forgot, i did that and simplified and i got my answer.

thank you very much for your patience :-)

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#20

(Original post by

i considered that also but i didnt know which would be best

**SashaNilssen**)i considered that also but i didnt know which would be best

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